An algebra problem...?

Question

I need help on this problem:

(-2b - 9)/(b2 + 7b + 12) = b/(b + 3) + 2/(b + 4)

I tried doing this, but it was no good. Help please?

(-2b - 9)/(b2 + 7b + 12) = b/(b + 3) + 2/(b + 4)
(-2b-9)/(b+3)(b+4) = b(b+4)/(b + 3)(b + 4) + 2(b+3)/(b + 3)(b + 4)
(-2b - 9)/(b + 3)(b + 4) = (b^2+4b+2b+6)/(b + 3)(b + 4)
(-2b - 9) = (b^2+6b+6)
0 = b^2 +8b + 15
0 = (b+5)(b-3)
b = -5, 3

Answer 1

Your least common denominator is (b+3)(b+4) = b^2 + 7b + 12, so multiply through by that. You'll get

-2b - 9 = b(b+4) + 2(b+3)

-2b - 9 = b^2 + 4b + 2b + 6

-2b - 9 = b^2 + 6b + 6

Now gather the terms together on one side:

b^2 +8b + 15 = 0

Solve this by any of the standard methods for solving quadratics. You'll get b = -5 or b = -3

Now remember: in your original equation, b = -3 and b = -4 both give 0 in the denominator, so they are not allowed as solutions. For that reason, we reject the -3 solution, so you're left with b = -5 as the only answer.

Answer 2

(-2b - 9)/(b2 + 7b + 12) = b/(b + 3) + 2/(b + 4)
(-2b-9)/(b+3)(b+4) = b(b+4)/(b + 3)(b + 4) + 2(b+3)/(b + 3)(b + 4)
(-2b - 9)/(b + 3)(b + 4) = (b^2+4b+2b+6)/(b + 3)(b + 4)
(-2b - 9) = (b^2+6b+6) given b not -3 or -4
0 = b^2 +8b + 15 given b not -3 or -4

0 = (b+5)(b+3)
b = -5 as -3 is not possible given b not -3 or -4.

Answer 3

(-2b - 9) / (b² + 7b + 12) = b/(b + 3) + 2/(b + 4)
(-2b-9)/(b+3)(b+4) = b(b+4)/(b + 3)(b + 4) + 2(b+3)/(b + 3)(b + 4)
(-2b - 9)/(b + 3)(b + 4) = (b²+4b+2b+6)/(b + 3)(b + 4)
(-2b - 9) = (b² + 6b + 6)_________Next line contain ur only fault)
b² +8b + 15 = 0
(b + 5) (b + 3) = 0
b = -5 , -3

Answer 4

(-2b-9) / (b^2 + 7b + 12) = b/(b+3) + 2(b+4)

(-2b-9) / ((b+3)(b+4)) = (b(b+4) + 2(b+3)) / (b+3)(b+4)

Set the numerators equal,
-2b-9 = b^2 +4b + 2b + 6 or
b^2 + 8*b + 15 = 0

solve given b can't equal -3 or -4

Answer 5

On your last factorization:

b2+8b + 15 should be (b+5)(b+3), so b = -5 or -3

Answer 6

3x-5x+x